If (4,2) and (k,-3) are conjugate points with respect to `x^(2)+y^(2)-5x+8y+6=0` them k=

Options

A. `28/3`
B. `-28/3`
C. `3/28`
D. `(-3)/28`

Answer

To find the value of \( k \) such that the points \( (4, 2) \) and \( (k, -3) \) are conjugate points with respect to the circle given by the equation \( x^2 + y^2 - 5x + 8y + 6 = 0 \), we will follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 - 5x + 8y + 6 = 0 \]
We can complete the square for \( x \) and \( y \).

#### Completing the square for \( x \):
\[ x^2 - 5x = (x - \frac{5}{2})^2 - \frac{25}{4} \]

#### Completing the square for \( y \):
\[ y^2 + 8y = (y + 4)^2 - 16 \]

Substituting these back into the equation gives:
\[ (x - \frac{5}{2})^2 - \frac{25}{4} + (y + 4)^2 - 16 + 6 = 0 \]
\[ (x - \frac{5}{2})^2 + (y + 4)^2 - \frac{25}{4} - 10 = 0 \]
\[ (x - \frac{5}{2})^2 + (y + 4)^2 = \frac{65}{4} \]

### Step 2: Find the Polar Equation
The polar of the point \( (4, 2) \) with respect to the circle can be derived using the formula:
\[ T = 0 \quad \text{where} \quad T = x_1 x + y_1 y - \frac{R^2}{4} \]
Here, \( (x_1, y_1) = (4, 2) \) and \( R^2 = \frac{65}{4} \).

Substituting these values:
\[ 4x + 2y - \frac{65}{4} = 0 \]

### Step 3: Simplify the Polar Equation
Multiplying through by 4 to eliminate the fraction:
\[ 16x + 8y - 65 = 0 \]

### Step 4: Substitute the Conjugate Point
Now, we substitute the conjugate point \( (k, -3) \) into this polar equation:
\[ 16k + 8(-3) - 65 = 0 \]
\[ 16k - 24 - 65 = 0 \]
\[ 16k - 89 = 0 \]

### Step 5: Solve for \( k \)
Now, solving for \( k \):
\[ 16k = 89 \]
\[ k = \frac{89}{16} \]

Thus, the value of \( k \) is \( \frac{89}{16} \).